Thermodynamics of Cells

1
Thermodynamics of Cells

• We have seen that welec q ?Ecell
• Also, recall that ?G wmax
• Therefore, ?G q ?Ecell
• Now, define 1 Faraday (F) of electrical charge as
  the charge in coulombs on 1 mole of electrons.
• 1 F (6.0221 x 1023 e/mol)(1.6022 x 1019 C/e)
• 96,485 C/mol e

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• If n the no. of moles of electrons gained or
  lost in the balanced cell reaction,
• then the total charge passed q n F
• And ?G n F ?E or ?G n F ?E
• Note that for a spontaneous reaction, ?G is
  negative and ?E is positive.
• Ex Zn(s) Zn2(1 M) Cu2(1 M) Cu(s)
• ?E 0.34 (0.76) 1.10 V.
• ?G (2 mol)(96485 C/mol)(1.10 V) 2.12 x 105
  J
• 212 kJ

3

• Kc exp(?G/RT) exp(2.12 x 105/8.314 x 298)
• e85.568 1.45 x 1037 strongly
  product-favored
• Using standard Gibbs free energies of formation
• ?G (1mol)(?Gf of Zn2) (1 mol)(?Gf of
  Cu2)
• (1 mol)(147.0 kJ/mol) (1mol)(65.5
  kJ/mol)
• 212.5 kJ
• Next, how can we relate ?E and Keq?
• ?G RT ln Keq and ?G n F ?E
• Therefore, RT ln Keq n F ?E

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• Rearrange to
• ln Keq (n F ?E) / (RT)
• Taking the antilog of both sides
• Keq exp (n F ?E) / (RT)
• At 298 K, RT/F 0.0257 V
• So, Keq exp(n ?E) / 0.0257 V at 298 K
• For Zn Zn2 Cu2 Cu
• Keq exp(2 1.10 V) / 0.0257 V) e85.603
• 1.50 x 1037

5

• Question What happens to ?E at equilibrium?
• Answer Since ?G n F ?E 0 at
  equili-brium, then ?E must be zero at
  equilibrium.
• 
  
• Now, lets look at cell potential when there are
  nonstandard conditions.
• Recall that ?G ?G RT lnQ
• and ?G n F ?E (or ?G n F
  ?E)
• We can substitute for the ?Gs in the first
  equation

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• n F ?E n F ?E RT ln Q
• Divide through by nF
• This is called the Nernst Equation.
• Or, at 298 K

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• Ex Calculate ?E for the following cell
• Zn(s) Zn2(0.100 M) Cu2(1.50 M) Cu(s)
• ?Ecell Ecath Ean 0.34 (0.76)
  1.10 V
• Zn(s) Cu2(aq) ? Zn2(aq) Cu(s)
• ?E 1.10 V (0.0348 V) 1.13 V
• The reaction is more spontaneous under these
  conditions, since there is less of the product
  and more of the reactant.

8

• Concentration Cells Use the same half-reaction
  at both electrodes but different concentrations.
• Ex Ag(s) Ag(0.10 M) Ag(1.0 M) Ag(s)
• Anode Ag(s) ? Ag(0.10 M) e
• Cathode Ag(1.0 M) e ? Ag(s)
• Overall Ag(1.0 M) ? Ag(0.10 M)
• ?Ecell 0.80 (0.80) 0.00 V
• Use the Nernst equation to calculate ?Ecell.

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Figure 17.9A Concentration Cell That Contains a
Silver Electrode and Aqueous Silver Nitrate in
Both Compartments
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• So, electrons are driven in the external circuit
  by a spontaneous dilution process.
• Ion-Specific Electrodes Used to measure the
  concentration of some species involved in a
  half-reaction.
• e.g., the glass pH electrode

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Figure 17.12A Glass Electrode for Measuring pH
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• Example of ion concentration measurements
• Pt(s) H2(g) H(aq) Cu2(1 M) Cu(s)
• Anode H2(g) ? 2 H(aq) 2 e E 0.00 V
• Cathode Cu2(aq) 2 e ? Cu(s) E 0.34 V
• Overall H2(g) Cu2(aq) ? 2 H(aq) Cu(s)
• ?E 0.34 0.00 0.34 V
• Set up the Nernst equation

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• Suppose that the measured cell potential is 0.50
  V.
• lnH2 (0.50 V 0.34 V)(2 / 0.0257 V)
• 12.451
• H2 e12.451 3.91 x 106
• H 1.98 x 103 M pH 2.70